Could someone explain these Algebra problems?

Question

1. A small company involved in a delivery business is in charge of two routes. A driver on route A travels 70 miles and a driver on route B travels 75 miles on a given day. The Driver on route A travels 5 mph slower than the driver on route B. Also the route A takes ½ an hour longer than B. How fast is each driver driving?

2. Two ferryboats leave opposite sides of a
lake at the same time. They pass each other when they are 800 meters from the nearest shore. When it reaches the oppostie side, each boat spends 30 minutes at the dock and then starts back. This time the boats pass each other when they are 400 meters from the nearest shore. Assuming that each of the boats travel at the same speed in both directions, how wide is the lake between the two ferry docks?

3. A train, an hour after starting, meets an accident which detains it one hour, after which it proceeds at 3/5 of its former rate and arrives 3 hours late. But, had an accident which happened 80 KM farther, it would have been 1.5 hours late only. Find the original rate of the train and the length of the journey.

Answer 1

1. Use the fact that distance=rate*time to set up two equations, one for driver A and one for driver B:

If x=speed of driver A and y= time of driver A,

70=x*y
75=(x+5)(y-.5)

Solve the first eqn for x or y (your choice): x=70/y

Plug this in to the other eqn: 75=(70/y+5)(y-.5)

Multiply this out: 75=70+5y-35/y-2.5

Multiply both sides by y: 75y=70y+5y^2-35-2.5y

Combine like terms: 0=5y^2-7.5y-35

Use the quadratic formula or factoring to find that y=3.5 and -2

Since a time of -2 doesn't make sense, we can eliminate that answer. So, the time for driver A is 3.5 hours. Put this into 70=x*y to get x=20. So, Driver A drives at 20 mph for 3.5 hours. Plugging these values into 75=(x+5)(y-.5) you find that Driver B drives 25mph for 3 hours.

2. No matter how I set this up I'm getting more variables than equations.

3. If we assume the train's regular speed is R, and use the fact that Dist=Rate*Time then the distance it covers is

R*1+0*1+3/5*R*T or R+3/5RT

If the accident had happend farther along, the distance would be the same as before but expressed this way:

R*1+80+0*1+3/5*R*(T-1.5) or R+80+3/5RT-9/5R

since the time for the second leg would be an hour and a half shorter this way than in the first scenario.

Setting them equal,

R+3/5RT=R+80+3/5RT-9/5R
0=80-9/5R
R=44.44...kph

Answer 2

Problem 1:
Already an excellent answer posted above (TychaBrahe)


Problem 2:
Ok, imagine the path across the river as a line joining the two docks A and B:

A--------------------------B
<---------W----------->

where the distance between the docks is W. Let the boat that leaves dock A (lets call it boat A) at the start have speed a, the boat that leaves dock B (call it boat B) at the start have speed b.
The distance X(A) of boat A from dock A after a time t is:
X(A)=at
The distance X(B) of boat B from dock A after the same time is:
X(B)=W-bt

At some moment t(1) they pass each other, i.e. their distances from dock A are the same:
at(1)=W-bt(1)
Rearranging gives:
W = (a+b)t(1)
or
t(1) = W/(a+b)

Now, let us assume that boat A is the slower boat, so that the nearest shore when the boats pass is the one that A started from; then
800 = at(1) = Wa/(a+b)

Now for the return journey (bear with me on this one...)
the distance of boat A from dock A at some time t (measured from the startof its journey from dock A, some considerable time ago now)
is
X(A) = W-a(t-(W/a+30))
[the term W/a is just the time it took A to cross the river on the outward journey; it waited 30 minutes then set out again; hence the total time it has been travelling on the return journey is t-(W/a+30)]
The distance of boat B from dock A is
X(B) = b(t-(W/b+30))

They pass at some time t(2):
b(t(2)-W/b-30) = W-a(t(2)-W/a-30)
expanding:
bt(2)-W-30b = W-at(2)+W+30a
rearranging:
3W+30(a+b) = (a+b)t(2)
finally:
t(2) = 30+3W/(a+b)
Now, we are assuming that boat A is the slower boat; hence it arrived at dock B later boat B did at dock A; hence it also left dock B later than boat B left dock A; therefore the nearest shore when they cross is that where dock B is;
therefore they cross when boat A is just 400 metres from dock B:
400 = a(t(2)-W/a-30)
(remember we are considering how far boat A is from dock B now)

400 = at(2)-W-30a = 30a+3aW/(a+b)-W-30a

from the outward journey we know that
800 = aW/(a+b)
so
400 = 30a+2400-W-30a
i.e.
W=2000


Partial answer to problem 3:

Let
A be the distance of the accident from the start of the journey in the first scenario.
J be the length of the journey in km
T be the usual duration of the journey in hours
V be the normal speed of the train kph

The actual duration of the journey consists of three components: the time taken to get to the accident; the wait; the time taken to complete the journey.
For the two scenarios these are:

scenario 1:
A/V+1+(J-A)*5/(3V) = T+3

scenario 2:
(A+80)/V + 1+(J-A-80)*5/(3V) = T+1.5

to simplify, multiply through by 3V in both cases:

S1:
3A+3V+5J-5A=3VT+9V

S2:
3A+240+3V+5J-5A-400 = 3VT+4.5V

subtract equation (2) from equation (1) to get:

160 = 4.5V
i.e.

V=320/9 = 35.555... kph

We are told that the train encounters the accident in scenario (1) after 1 hour of travel, hence A and V are numerically equal. This enables us to re-write the above scenarios:

S1:
5J = 3VT + 8V

S2:
5J = 160 + 3VT + 3.5V


where the value of V is derived above; however, the two equations are identical, so I don't see a way of proceeding to a complete solution from here.

Answer 3

Lets say that B takes t time to finish the route. Then B's speed is 75/t mph. A's speed is then 70/(t +0.5) mph

75/t - 70/(t + 0.5) = 5

Let's get some common denominators.

75(t + 0.5)/(t^2 + 0.5t) - 70(t)/(t^2 + 0.5t) = 5(t^2 + 0.5t)/(t^2 + 0.5t)

So (75t + 37.5) - 70t = 5t^2 + 2.5t
5t + 37.5 = 5t^2 + 2.5t
5t^2 - 2.5t - 37.5 = 0

Let's make that easier to read.

10t^2 - 5t - 75 = 0
2t^2 - t - 15 = 0

(2t + 5)(t - 3)

So t = 3 or t = -5/2. Well, t can't be a negative, so t = 3.

75 / 3 = 25 mph for truck B

70 / (3 + .5) = 20 mph for truck A

And there's your 5 mph difference.




This doesn't make any sense.

1. The boats are traveling the same speed.
2. The boats leave at the same time.
3. The boats are traveling opposite directions on the same route.

By definition, after some time t, which depends on the width of the lake and their speed, the boats should meet in the middle of the lake. They should arrive at the opposite shores at the same time. Since they delay and leave again at the same time, they will meet again in the middle of the lake. There's no way that they could be 800 m from the shore at one meeting and 400 m from the shore on the other.


Ack! I've tried this problem twice and gotten stuck both times!
Train normally runs distance D in time T at a speed S = D / T.
Train is traveling at S kph for one hour.
Train is delayed for one hour.
Train continues at 3S/5 for t hours over D-x = d distance.
Train is three hours late.
Train took two hours longer to run the route (lateness - delay).
So, (3S/5) / (d) = (T - 1) + 2

(The reduced speed divided by the remaining distance = the normal time minus the one hour run at the usual rate plus the two hours it took longer than normal.)

On the other hand, had the accident happened 80 km further along, (3S/5) / (d-80)) = (T - 1) + .5

((3S/5) / (d) - 2 = (T - 1)
((3S/5) / (d-80)) - .5 = (T - 1)

So ((3S/5) / (d)) - 2 = ((3S/5) / (d-80)) - .5
((3S/5) / (d)) - ((3S/5) / (d-80)) - 1.5 = 0

[I think I hate your math teacher.]

(3S/5)(d-80) / (d)(d-80) - (3S/5)(d) / (d)(d-80) - 1.5(d)(d-80) / (d)(d-80) = 0

(3S/5)(d-80) - (3S/5)(d) - 1.5(d)(d-80) = 0
(3S)(d-80) - (3S)(d) - 7.5(d)(d-80) = 0
(6S)(d-80) - (6S)(d) - 15(d)(d-80) = 0
(2S)(d-80) - (2S)(d) - 5(d)(d-80) = 0
2Sd - 160S - 2Sd - 5d^2 + 40d = 0
5d^2 - 40d + 160S = 0

160S = 5d^2 - 40d

Answer 4

since the others are having trouble with it, I'll work on 2
we can assume that one boat is faster than the other (each one travels the same speed in both directions, but difff from each other). We'll call one Fast and one Slow.
We'll also say that Fast leaves dock A first and Slow leaves dock B, and the distance accross the lake is d.
we can assume that the 'nearest shore' on the first leg is dock B, since the Fast boat crossed more lake.
likewise the 'nearest shore' on the second leg is dock A since the Fast boat is well ahead.
Sketch that out...
we can set up an equation going between the two interceptions.
for the fast boat distance traveled is 800m+d-400=d+400.
for the slow boat distance traveled is d-800+400=d-400
since time taveled is the same between the points of interception, and speed = distance/time, we can set up a ratio of the speeds
Fast/Slow = [(d+400)/T] / [(d-400)/T] = (d+400)/(d-400)
From the very first part of the trip we know that
800/Slow = (d-800)/Fast
Fast/Slow = (d-800)/800
so
(d+400)/(d-400) = (d-800)/800
solve for d

Answer 5

it potential (d-4) 9 cases is the comparable factor as 5d + 8 you may remedy it with the help of multiplying d-4 with the help of 9 which makes it 9d-36 then you definately can subtract 5d from the two aspects because of the fact what you do to a minimum of one edge must be achieved to the different to maintain each and every thing equivalent. that ought to bypass away 4d-36 = +8 then you definately can upload 36 to the two aspects to get rid of the -36 (because of the fact -36 plus 36 equals 0 precise?) with the help of the 4d yet nonetheless save it interior the equation so that is nonetheless the comparable answer. if it particularly is clever now your equation is 4d = 8 + 36 take the 8 + 36 sum and divide it with the help of four (do the comparable factor to the different edge so your left with basically d) in case you divide the two aspects with the help of four you wont have what 4d equals you all have what d equals precise? you ought to get d = 11 seeing it multi functional bypass helps me on occasion so it would appear like this: 9( d - 4 ) = 5d + 8 9d - 36 = 5d + 8 4d - 36 = 8 4d = 8 + 36 4d = 40 4 d = 11 desire this wasn't too difficult, that is lots extra straightforward to describe in individual positioned on i will factor and draw arrows and clarify in extra intensity. asking your instructor may well be a competent thought - or somebody you already know who's good at considered one of those factor and could permit you already know approximately it.